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For programming bipolar ROMs I need a programming voltage of U=27V,30V,33V and an input voltage of U-7V. My first idea was to use a Zener diode in series to drop 7V at the diode. However, the needed current of at least 230 mA is too high for usual Zener diodes.

Is there a simple way to get the 2nd voltage with a transistor?

schematic

simulate this circuit – Schematic created using CircuitLab

Might it also work with a negative voltage regulator like the 7905, with a zener somehow to generate -7V with respect to the U rail and then use this negative voltage but with respect to the real ground?

schematic

simulate this circuit

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  • \$\begingroup\$ In the 2nd picture, it is of course not 1V but just U \$\endgroup\$ – Andi Sägezahntiger Jul 2 at 23:01
  • \$\begingroup\$ That's easy tinyurl.com/ya2wpyag \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 2 at 23:12
  • \$\begingroup\$ Which bipolar ROMs? What voltage tolerance is allowed? \$\endgroup\$ – Bruce Abbott Jul 3 at 0:11
  • \$\begingroup\$ @BruceAbbott MMI 6301, 256x4 bit, the LSI bipolar data book does say nothing about a tolerance except the range must be 20V to 26V. First try 20V three times, if it didnt blow the fuse, try 23V tree times, and at last 26 volts three times... So since I only have 5 of these chips and two must be programmed correctly I would say 26V is the upper limit without tolerance... \$\endgroup\$ – Andi Sägezahntiger Jul 3 at 0:28
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 Thanks! \$\endgroup\$ – Andi Sägezahntiger Jul 3 at 0:33
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You could try something like this, though the accuracy might not be good enough.

schematic

simulate this circuit – Schematic created using CircuitLab

Q1 dissipates about 7V * 256mA = 1.8W so it needs a bit of a heat sink.

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